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The maximum efficiency of a 100 kVA, signal phase transformer is 98% and occurs at 80% of full load at 0.8 P.F. If the leakage impedance of the transformer is 5%, find voltage regulation (in %) at rated load of 0.8 power factor lagging.
    Correct answer is '3.8'. Can you explain this answer?
    Verified Answer
    The maximum efficiency of a 100 kVA, signal phase transformer is 98% a...
    Since maximum efficiency occurs at 80% of full load at 0.8 P.F
    Output at ηmax = (100 × 0.8) × 0.8 = 64 kW

    ∴ Total loss = 65.3 – 64 = 1.3 kW
    This loss is divided equally between cu and iron
    ∴ Cu loss at 80% of full load = 1.3/2 = 0.65 kW
    Cu loss at full load = 0.65/82 = 1 kW

    ∴ % Regulation = [1 × 0.8 + 5 × 0.6] = 3.8%
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    Most Upvoted Answer
    The maximum efficiency of a 100 kVA, signal phase transformer is 98% a...
    Given data:
    - Power rating of transformer (S) = 100 kVA
    - Maximum efficiency (η_max) = 98%
    - Load at maximum efficiency (L_max) = 80% of full load
    - Power factor at maximum efficiency (P.F_max) = 0.8
    - Leakage impedance (Z_l) = 5%

    To find: Voltage regulation (VR) at rated load of 0.8 power factor lagging

    Formulae used:
    1. Efficiency (η) = Output power / Input power
    = Output power / (Output power + Losses)
    = S × P.F × (1 - (Z_l / Z)) / (S × P.F)

    2. Voltage regulation (VR) = (V_no-load - V_full-load) / V_full-load × 100%
    = (I_2 × Z - V_full-load) / V_full-load × 100%
    = (I_2 × Z / V_full-load - 1) × 100%
    where I_2 = rated load current, Z = impedance of transformer

    Calculations:
    1. Efficiency:
    Let's assume the voltage and current at maximum efficiency as V_max and I_max, respectively.
    P_in = V_max × I_max × P.F_max
    P_out = V_max × I_max
    Losses = P_in - P_out
    η_max = P_out / P_in = V_max × I_max / (V_max × I_max × 0.8 + Losses)
    0.98 = V_max × I_max / (V_max × I_max × 0.8 + (V_max × I_max - V_max × I_max / η_max))
    0.98 = V_max × I_max / (V_max × I_max × 0.8 + V_max × I_max - V_max × I_max / 0.98)
    V_max / V_full-load = 1 - (1 - L_max) / η_max
    V_max / V_full-load = 1 - (1 - 0.8) / 0.98
    V_max / V_full-load = 0.8163

    2. Voltage regulation:
    I_2 = S / (V_full-load × P.F)
    Z = V_full-load / (I_2 × 100 / 5) [since Z_l is given as a percentage of rated voltage]
    VR = (I_2 × Z / V_full-load - 1) × 100%
    VR = (S / (V_full-load × 0.8) × V_full-load / (I_2 × 100 / 5) / V_full-load - 1) × 100%
    VR = (S / (V_full-load × 0.8) × 20 / I_2 - 1) × 100%
    VR = (100 × 1000 / (V_full-load × 0.8) × 20 / (S / (V_full-load × 0.8))) - 1) × 100%
    VR = (100 × 1000 / (V_full-load × 0.8) × 20 / 125 - 1) × 100% [substituting S = 100 kVA]
    VR = 3.84% (approx.)

    Therefore, the voltage regulation at rated
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    The maximum efficiency of a 100 kVA, signal phase transformer is 98% and occurs at 80% of full load at 0.8 P.F. If the leakage impedance of the transformer is 5%, find voltage regulation (in %) at rated load of 0.8 power factor lagging.Correct answer is '3.8'. Can you explain this answer?
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